Analysis questions:
#1: Calculate the enthalpy for this reaction:
2C(s) + H2(g) —> C2H2(g) ΔH° = ??? kJ
Given the following thermochemical equations:
C2H2(g) + 5⁄2O2(g) —> 2CO2(g) + H2O(ℓ) ΔH° = −1299.5 kJ
C(s) + O2(g) —> CO2(g) ΔH° = −393.5 kJ
H2(g) + 1⁄2O2(g) —> H2O(ℓ) ΔH° = −285.8 kJ
ANSWER: +226.7. kJ
#2: Calculate the enthalpy of the following chemical reaction:
CS2(ℓ) + 3O2(g) —> CO2(g) + 2SO2(g)
Given the following thermochemical equations:
C(s) + O2(g) —> CO2(g) ΔH = −393.5 kJ/mol
S(s) + O2(g) —> SO2(g) ΔH = −296.8 kJ/mol
C(s) + 2S(s) —> CS2(ℓ) ΔH = +87.9 kJ/mol
ANSWER: 1075 kJ
#3: Given the following data:
SrO(s) + CO2(g) —> SrCO3(s) ΔH = −234 kJ
2SrO(s) —> 2Sr(s) + O2(g) ΔH = +1184 kJ
2SrCO3(s) —> 2Sr(s) + 2C(s, gr) + 3O2(g)ΔH = +2440 kJ
Find the ΔH of the following reaction:
C(s, gr) + O2(g) —> CO2(g)
ANSWERl -394 kJ
#4: Using the following thermochemical equations, calculate the standard enthalpy of combustion for one mole of liquid acetone (C3H6O).
3C(s) + 3H2(g) + 1⁄2O2(g) —> C3H6O(ℓ) ΔH° = −285.0 kJ
C(s) + O2(g) —> CO2(g) ΔH° = −394.0 kJ
H2(g) + 1⁄2O2(g) —> H2O(ℓ) ΔH° = −286.0 kJ
ANSWER: -1755 kJ
#5 The standard enthalpy change of formation of propane is impossible to measure directly. That is because carbon and hydrogen will not directly react to make propane. However, standard enthalpy changes of combustion are relatively easy to measure.
C3H8(g) ΔH1 = −2219.9 kJ
C(s, gr) ΔH2 = −393.5 kJ
H2(g) ΔH3 = −285.8 kJ
Answer: -103.8 kJ